Adding Class Name in JSON using C#

Question

I created below JSON of Person class using JSON.NET But "Person" is not showing up anywhere in JSON. I think it should show up in the start. What is the problem or how to resolve? Thank you.

[
  {
    "Name": "Umer",
    "Age": 25
  },
  {
    "Name": "Faisal",
    "Age": 24
  }
]

C# code is here which generated JSON

List<Person> eList = new List<Person>();
Person a = new Person("Umer",25);
Person b = new Person("Faisal", 24);
eList.Add(a);
eList.Add(b);
string jsonString = JsonConvert.SerializeObject(eList,Formatting.Indented);

This might help: http://stackoverflow.com/questions/10144300/json-net-class-name-as-root — Jeremy Thompson, Jul 04 '15 at 11:12

kyrylomyr · Answer 1 · 2015-07-04T17:20:29.177

You need to add a TypeNameHandling setting:

List<Person> eList = new List<Person>();
Person a = new Person("Umer", 25);
Person b = new Person("Faisal", 24);
eList.Add(a);
eList.Add(b);
string jsonString = JsonConvert.SerializeObject(eList, Formatting.Indented,
    new JsonSerializerSettings { TypeNameHandling = TypeNameHandling.All });

This way each JSON object will have an additional field "$type":

[
    {
        "$type" : "YourAssembly.Person"
        "Name" : "Umer",
        "Age" : 25
    },
    ...
]

For more details see the documentation.

score 1 · Answer 2 · edited May 23 '17 at 12:24

1

There is no problem in that.It can be deserialized that way.

You can deserialize it like that :

Person deserialized = (Person)JsonConvert.DeserializeObject( serializedText ,typeof(Person))

But if you need the root this question may help.

edited May 23 '17 at 12:24

Community

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answered Jul 04 '15 at 11:08

TC Alper Tokcan

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He's not asking about deserialisation, from what I gather he wants the root of the json to start with the word Person – Jeremy Thompson Jul 04 '15 at 11:10

msmolcic · Answer 3 · 2015-07-04T12:22:55.500

1

You could use anonymous class to recreate your list and then serialize it if you really want class name as part of your JSON.

var persons = eList.Select(p => new { Person = p }).ToList();
var json = JsonConvert.SerializeObject(persons, Formatting.Indented);

Output:

JSON image

edited Jul 04 '15 at 12:22

answered Jul 04 '15 at 11:25

msmolcic

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Wouldn't just `eList.Select(p => new { Person = p })` suffice? We know that each person object is serialized properly. We just need to key each object to the string "Person". Please correct me if I missed something. – Amith George Jul 04 '15 at 12:16
@AmithGeorge You're absolutely right. I edited my answer, thanks. – msmolcic Jul 04 '15 at 12:23

score 1 · Answer 4 · edited Jul 04 '15 at 12:12

1

Try

var Person = new List<Person>();
Person a = new Person("Umer", 25);
Person b = new Person("Faisal", 24);
Person.Add(a);
Person.Add(b);

var collection = Person;
dynamic collectionWrapper = new {
  myRoot = collection
};

var output = JsonConvert.SerializeObject(collectionWrapper);

edited Jul 04 '15 at 12:12

Amith George

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answered Jul 04 '15 at 11:44

Sreejith

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in the line `var collection = Person`, did you mean to instead type `"Person"`, ie a string? – Amith George Jul 04 '15 at 12:13

Adding Class Name in JSON using C#

4 Answers4